CIRCLES C3

In the figure below, find the value of <BDA.

Solution

<ABD = <DAC

∴ <ABD = 51 [angles in alternate segments are equal.]

               Consider ∆ABD

<ABD + <BAD+ <BDA = 180⁰ [total degrees of a triangle]

51 + (60 + 51) + <BDA= 180⁰

51 + 111 + <BDA= 180⁰

162 + <BDA= 180⁰

<BDA= 180⁰ -162

<BDA= 18⁰  

∴ <BDA= 18⁰